$g(x)=2x^2+3x$ What is the average rate of change of $g$ over the interval $[-2,-2+h]$, in terms of $h$, where $h\neq 0$ ? Your answer must be fully expanded and simplified.
Answer: This is the formula for the average rate of change of a function $f$ over the interval $[a,b]$ : $\dfrac{f(b)-f(a)}{b-a}$ We can calculate that $g(-2)=2$. We are interested in the average rate of change of $g(x)=2x^2+3x$ over the interval $[-2,-2+h]$ : $\begin{aligned} &\phantom{=}\dfrac{g(-2+h)-g(-2)}{(-2+h)-(-2)} \\\\ &=\dfrac{2(-2+h)^2+3(-2+h)-2}{-2+h-(-2)} \\\\ &=\dfrac{2(-2+h)^2+3(-2+h)-2}{h} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} &\phantom{=}\dfrac{2(-2+h)^2+3(-2+h)-2}{h} \\\\ &=\dfrac{2(4-4h+h^2)+3(-2+h)-2}{h} \\\\ &=\dfrac{8-8h+2h^2-6+3h-2}{h} \\\\ &=\dfrac{2h^2-5h}{h} \\\\ &=\dfrac{h(2h-5)}{h} \\\\ &=2h-5\text{, for }h\neq 0 \end{aligned}$ Since we are given that $h\neq 0$, the average rate of change of the function is $2h-5$. Notice that the average rate of change is calculated just like the slope of the secant line that intersects the graph of the function at the interval's endpoints.